# Radius of a circle

**The radius of a circle**, a fundamental geometric concept, represents the distance from the circle’s center to any point on its boundary. The equation to find it is

**r = C/(2π)**, where C is the circumference.

## Problem 1: Circumference and Radius

**Problem: The circumference of a circular garden is 150 meters. Find the radius of the garden, and also calculate the area enclosed by the garden.**

**Solution:**

Circumference (C) = 2πr, where r is the radius.

Given C = 150 meters, solve for r: 150 = 2πr, r = 150 / (2π) ≈ 23.81 meters (rounded to two decimal places).

Area (A) = πr² = π * (23.81)² ≈ 1785.47 square meters (rounded to two decimal places).

## Problem 2: Concentric Circles

**Problem: Two concentric circles have radii of 10 cm and 15 cm, respectively. Calculate the area between the two circles.**

**Solution:**

Area between the circles = Area of the larger circle – Area of the smaller circle.

Area = π * (15)² – π * (10)² = π * (225 – 100) = π * 125 ≈ 392.7 square cm (rounded to one decimal place).

## Problem 3: Tangent Line to a Circle

**Problem: A line is tangent to a circle with a radius of 6 inches. If the distance from the center of the circle to the point where the line touches the circle is 3 inches, find the length of the line.**

**Solution:**

Using the Pythagorean theorem, the length of the line is the hypotenuse of a right triangle.

Hypotenuse = √(radius² + distance from center²) = √(6² + 3²) = √(36 + 9) = √45 ≈ 6.71 inches (rounded to two decimal places).

## Problem 4: Sector of a Circle

**Problem: A sector of a circle with a radius of 12 meters has a central angle of 60 degrees. Find the area of the sector.**

**Solution:**

Area of the sector = (θ/360) * πr², where θ is the central angle and r is the radius.

Area = (60/360) * π * (12)² = (1/6) * π * 144 = 24π square meters.

## Problem 5: Chord Length

**Problem: In a circle with a radius of 8 cm, a chord is drawn at a distance of 6 cm from the center of the circle. Find the length of the chord.**

**Solution:**

The chord divides the circle into two equal parts.

Using the Pythagorean theorem, the length of the chord = 2 * √(radius² – distance from center²) = 2 * √(8² – 6²) = 2 * √(64 – 36) = 2 * √28 ≈ 10.58 cm (rounded to two decimal places).

## Problem 6: Area of Inscribed Circle

**Problem: A circle with a radius of 5 cm is inscribed in a square. Find the area of the square.**

**Solution:**

The diameter of the inscribed circle is equal to the side length of the square.

Diameter = 2 * 5 = 10 cm.

The side length of the square = Diameter = 10 cm.

Area of the square = side length² = 10² = 100 square cm.

## Problem 7: Sector Angle and Arc Length

**Problem: In a circle with a radius of 14 cm, an arc has a length of 22 cm. Find the central angle in degrees subtended by this arc.**

**Solution:**

Arc length (L) = (θ/360) * 2πr, where θ is the central angle and r is the radius.

22 = (θ/360) * 2π * 14.

Solve for θ: θ = (22 * 360) / (2π * 14) ≈ 90 degrees (rounded to the nearest degree).

## Problem 8: Segment of a Circle

**Problem: A sector of a circle with a radius of 9 cm and a central angle of 120 degrees is cut out. Find the area of the remaining segment.**

**Solution:**

Area of the segment = Area of the sector – Area of the triangle formed by the central angle.

Area = ((120/360) * π * (9)²) – (1/2 * 9 * 9 * sin(120°)).

Area ≈ (1/3) * π * 81 – (1/2 * 9 * 9 * √3/2) ≈ 28.27 square cm (rounded to two decimal places).

## Problem 9: Inscribed Angle

**Problem: In a circle with a radius of 10 cm, an inscribed angle measures 45 degrees. Find the length of the intercepted arc.**

**Solution:**

The intercepted arc length = (θ/360) * 2πr, where θ is the inscribed angle and r is the radius.

Arc length = (45/360) * 2π * 10 = (1/8) * 20π = 5π cm.

## Problem 10: Area of a Sector

**Problem: A sector of a circle with a radius of 7 cm has an area of 14 square cm. Find the central angle in degrees subtended by this sector.**

**Solution:**

Area of the sector = (θ/360) * πr².

14 = (θ/360) * π * 7².

Solve for θ: θ = (14 * 360) / (π * 7²) ≈ 120 degrees (rounded to the nearest degree).

These lengthy problems involving the radius of a circle cover various aspects, including circumference, area, tangents, sectors, chords, and more.

Problem 11: Circle Inscribed in a Rectangle

**Problem: A rectangle has dimensions of 10 cm by 15 cm. A circle is inscribed inside the rectangle such that it touches all four sides. Calculate the radius of the inscribed circle and the area of the shaded region outside the circle but within the rectangle.**

**Solution:**

To find the radius of the inscribed circle (r), we can use the fact that it touches all four sides of the rectangle. The radius is half the length of the shorter side of the rectangle. So, r = 10 cm / 2 = 5 cm.

The area of the shaded region is the difference between the area of the rectangle and the area of the inscribed circle.

Area of rectangle = length * width = 10 cm * 15 cm = 150 square cm.

Area of inscribed circle = πr² = π * (5 cm)² = 25π square cm.

Therefore, the area of the shaded region = 150 square cm – 25π square cm ≈ 68.54 square cm (rounded to two decimal places).

Problem 12: Circle Tangent to Two Lines

**Problem: Two lines with equations 2x + 3y = 6 and x – y = 4 intersect at point P. A circle is tangent to both lines and has its center at point O. Find the coordinates of point O.**

**Solution:**

To find the coordinates of point O, we first need to find the coordinates of point P, where the two lines intersect. We can solve the system of equations:

2x + 3y = 6

x – y = 4

Solve the second equation for x: x = 4 + y.

Substitute this expression for x into the first equation: 2(4 + y) + 3y = 6.

Simplify and solve for y: 8 + 2y + 3y = 6, 5y = -2, y = -2/5.

Substitute the value of y back into the second equation to find x: x – (-2/5) = 4, x = 4 + 2/5, x = 22/5.

So, the coordinates of point P are (22/5, -2/5).

The center of the circle, point O, will be equidistant from both lines. It will also be on the line perpendicular to the lines 2x + 3y = 6 and x – y = 4 and passing through the midpoint of the segment joining point P to the point where the two lines intersect.

Calculate the midpoint of the segment joining (22/5, -2/5) and the intersection point (22/5, -2/5): [(22/5 + 22/5)/2, (-2/5 – 2/5)/2] = (44/10, -4/10) = (22/5, -2/5).

The line perpendicular to 2x + 3y = 6 has a slope of -3/2, and its equation is y – (-2/5) = (-3/2)(x – 22/5).

Simplify the equation to find the equation of the line passing through O and perpendicular to both lines: y + 2/5 = (-3/2)(x – 22/5).

Now, find the intersection of this line with either of the original lines to get the coordinates of point O.

Solving the equation -3/2(x – 22/5) = 2x + 3y = 6 simultaneously gives x = 22/3 and y = -2/3.

So, the coordinates of point O are (22/3, -2/3).

Problem 13: Concentric Circles and Area Ratio

**Problem: A small circle with a radius of 5 cm is drawn inside a larger circle with a radius of 10 cm, with both circles having the same center. Calculate the ratio of the areas of the smaller circle to the larger circle.**

**Solution:**

The ratio of the areas of two circles is equal to the ratio of the squares of their radii.

Area of the smaller circle = π * (5 cm)² = 25π square cm.

Area of the larger circle = π * (10 cm)² = 100π square cm.

Calculate the ratio of their areas: (25π) / (100π) = 1/4.

So, the ratio of the areas of the smaller circle to the larger circle is 1:4.

## Problem 14: Circle and Tangent Line

**Problem: A circle with a radius of 6 cm is centered at point O. A tangent line to the circle is drawn from an external point P, which is 8 cm away from point O. Calculate the length of the tangent line segment from point P to the circle.**

**Solution:**

Draw a line segment from point O to point P, which is the radius of the circle. This creates a right triangle with the tangent line as the hypotenuse.

The length of the radius (OA) is 6 cm, and the length of OP is 8 cm.

Using the Pythagorean theorem, we can find the length of the tangent line (TP):

TP² = OP² – OA²

TP² = (8 cm)² – (6 cm)²

TP² = 64 cm² – 36 cm²

TP² = 28 cm².

Taking the square root of both sides to find TP: TP = √28 cm ≈ 5.29 cm (rounded to two decimal places).

So, the length of the tangent line segment from point P to the circle is approximately 5.29 cm.

## Problem 15: Circle Inscribed in an Equilateral Triangle

**Problem: An equilateral triangle with sides of length 18 cm has a circle inscribed within it. Find the radius of the inscribed circle and the area of the shaded region between the circle and the triangle.**

**Solution:**

In an equilateral triangle, the radius of the inscribed circle is one-third of the height of the triangle.

To find the height of the equilateral triangle, we can use the formula: height = (√3/2) * side length.

Height = (√3/2) * 18 cm = 9√3 cm.

The radius of the inscribed circle is one-third of the height: r = (1/3) * 9√3 cm = 3√3 cm.

To find the area of the shaded region, we need to subtract the area of the inscribed circle from the area of the equilateral triangle.

Area of the inscribed circle = πr² = π * (3√3)² = 9π square cm.

Area of the equilateral triangle = (1/2) * base * height = (1/2) * 18 cm * 9√3 cm = 81√3 square cm.

Area of the shaded region = Area of the equilateral triangle – Area of the inscribed circle = 81√3 square cm – 9π square cm ≈ 112.35 square cm (rounded to two decimal places).

These extensive problems involving circles cover topics such as inscribed circles, tangents, concentric circles, and more, and provide detailed solutions to each problem.

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